In ∆ABC, m∠ACB = 90°, m∠A = 40°, and D ∈ AB such that CD is perpendicular to side AB. Find m∠DBC and m∠BCD.

 ∠B = 50°

               ∠BCD = 40°

Step-by-step explanation:

ACB is a right triangle where ∠A = 40° and ∠C = 90°.

Use the Triangle Sum Theorem for ΔABC to find ∠B:

∠A + ∠B + ∠C = 180°

40° + ∠B + 90° = 180°

         ∠B + 130° = 180°

                  ∠B = 50°

BCD is a right triangle where ∠B = 50° and ∠D = 90°.

Use the Triangle Sum Theorem for ΔBCD to find ∠C:

∠B + ∠C + ∠D = 180°

50° + ∠C + 90° = 180°

         ∠C + 140° = 180°

                  ∠C = 40°

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

2 x t divided by n

step-by-step explanation: